Question

Calculate emf of the following cell at 298 K :
\( \text{Zn (s)| Zn}^{2+}\text{(aq) (0.1 M)|| Ag}^+\text{ (aq) (0.01 M)|Ag (s)} \)
(Given : \( E^\circ_{\text{Zn}^{2+}/\text{Zn}} = -0.76 \text{ V}, E^\circ_{\text{Ag}^{+}/\text{Ag}} = +0.80 \text{ V, [log 10 = 1]} \))

Hint:

Always square the concentration of \( \text{Ag}^+ \) in the log term because of the stoichiometric coefficient 2.
Remember \( \log(10^x) = x \).
A positive \( E_{\text{cell}} \) indicates a spontaneous reaction.

Answer 1

Step 1: Write the cell reaction and find the number of electrons.
The cell notation $\text{Zn(s)} \mid \text{Zn}^{2+}(0.1\ M) \parallel \text{Ag}^+(0.01\ M) \mid \text{Ag(s)}$ tells us zinc is oxidised at the anode and silver ion is reduced at the cathode. \[ \text{Zn} \rightarrow \text{Zn}^{2+} + 2e^-, \quad 2\text{Ag}^+ + 2e^- \rightarrow 2\text{Ag} \] Combining these, the overall reaction transfers $n = 2$ electrons.
Step 2: Find the standard cell potential.
\[ E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} = 0.80 - (-0.76) = 1.56\ \text{V} \]
Step 3: Simplify the concentration ratio before applying the Nernst equation.
The reaction quotient for this cell is $Q = \dfrac{[\text{Zn}^{2+}]}{[\text{Ag}^+]^2}$. Putting in the numbers first: \[ Q = \frac{0.1}{(0.01)^2} = \frac{10^{-1}}{10^{-4}} = 10^{3} \]
Step 4: Apply the Nernst equation and solve for $E_{cell}$.
\[ E_{cell} = E^{\circ}_{cell} - \frac{0.059}{n}\log Q = 1.56 - \frac{0.059}{2}\log(10^{3}) \] Since $\log(10^3) = 3$, \[ E_{cell} = 1.56 - (0.0295 \times 3) = 1.56 - 0.0885 = 1.4715\ \text{V} \]
\[ \boxed{E_{cell} = 1.4715\ \text{V}} \]